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2 years ago

In an experiment 25.0 mL of 0.100 M KI was diluted to 50.0 mL. Calculate the molarity of the diluted solution

Chemistry

1 answer:

Anna11 [10]2 years ago

3 0

Answer:

The molarity is "0.050 M".

Explanation:

The given values are:

M1 = 0.100 M

M2 = ?

V1 = 25.0 mL

V2 = 50.0 mL

As we know,

⇒ $M1\times V1=M2\times V2$

Or,

⇒ $M2=\frac{M1\times V1}{V2}$

By putting the values, we get

$=\frac{0.100\times 25}{50}$

$=\frac{2.5}{50}$

$=0.05 \ M$

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An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective

djverab [1.8K]

Answer:

The new partial pressures after equilibrium is reestablished for $PCl_3$ :

$P_1'=6.798 Torr$

The new partial pressures after equilibrium is reestablished $Cl_2$ :

$P_2'=26.398 Torr$

The new partial pressures after equilibrium is reestablished for $PCl_5$ :

$P_3'=223.402 Torr$

Explanation:

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At equilibrium before adding chlorine gas:

Partial pressure of the $PCl_3=P_1=13.2 Torr$

Partial pressure of the $Cl_2=P_2=13.2 Torr$

Partial pressure of the $PCl_5=P_3=217.0 Torr$

The expression of an equilibrium constant is given by :

$K_p=\frac{P_1}{P_1\times P_2}$

$=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245$

At equilibrium after adding chlorine gas:

Partial pressure of the $PCl_3=P_1'=13.2 Torr$

Partial pressure of the $Cl_2=P_2'=?$

Partial pressure of the $PCl_5=P_3'=217.0 Torr$

Total pressure of the system = P = 263.0 Torr

$P=P_1'+P_2'+P_3'$

$263.0Torr=13.2 Torr+P_2'+217.0 Torr$

$P_2'=32.8 Torr$

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At initail

(13.2) Torr (32.8) Torr (13.2) Torr

At equilbriumm

(13.2-x) Torr (32.8-x) Torr (217.0+x) Torr

$K_p=\frac{P_3'}{P_1'\times P_2'}$

$1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}$

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished for $PCl_3$ :

$P_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr$

The new partial pressures after equilibrium is reestablished $Cl_2$ :

$P_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr$

The new partial pressures after equilibrium is reestablished for $PCl_5$ :

$P_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr$

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